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Nina [5.8K]
4 years ago
7

One of the pieces of evidence supporting energy quantization was the line spectra of elements. Why does this demonstrate energy

quantization?
a) Photons are packets of energy that act both as a particle and a wave.
b) The spectra is continuous and shows emission at all wavelengths.
c) There are sharp emission lines demonstrating discrete energy levels.
d) All compounds emit light in the ultraviolet and visible regions specifically.​
Chemistry
1 answer:
bezimeni [28]4 years ago
8 0

Answer:

c) There are sharp emission lines demonstrating discrete energy levels.

Explanation:

When an element emits energy in the form of radiation, it produces a spectrum of colors on a photographic plate. This spectrum can either be continuous or discrete. In continuous spectrum the spectrum continues without any discrimination between two regions. This represents the continuous emission of radiation, and thus the continuous emission of energy without any break.

On the other hand, the line spectrum consists of discrete and sharp lines, which shows the emission of radiation in a certain amount in a certain time, with a break between emission. Hence, the line spectra supports the quantization of energy.

The correct option is:

<u>c) There are sharp emission lines demonstrating discrete energy levels.</u>

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What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10
vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

<h3>What are pH and pOH?</h3>

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, \rm pOH = 14 - pH

The pH of KOH can be calculated by the formula,

\rm pH = \rm -log [H^{+}]

In the first case, the concentration of the KOH is 1. 87 \times  10^{-13}\;\rm  M

Substituting values in the equation:

\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times  10^{-13}\;\rm  M ]\\\\&= 12.73\end{aligned}

Hence, the pH of KOH is 12.73.

<u />

pOH of KOH can be calculated by the formula,

\rm pOH = \rm -log [OH^{-}]

The hydroxide concentration of the KOH solution is 5. 81 \times 10^{-3}\;\rm  M

Substituting value in the equation:

\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm  M ]\\\\&= 2.24 \end{aligned}

Hence, the pOH of KOH is 2.24

<u />

The pH of NaCl can be calculated by the formula,

\rm pH = \rm -log [H^{+}]

In the third case, the concentration of the NaCl is 1. 00\times 10^{-7}\;\rm  M

Substituting values in the equation:

\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times  10^{-7}\;\rm  M ]\\\\&= 7 \end{aligned}

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

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